package com.sxkiler.demo.easy;

import org.junit.jupiter.api.Assertions;
import org.junit.jupiter.api.Test;
import java.util.*;
import com.sxkiler.demo.model.*;

/**
range-sum-query-immutable=区域和检索 - 数组不可变
<p>给定一个整数数组 &nbsp;<em>nums</em>，求出数组从索引&nbsp;<em>i&nbsp;</em>到&nbsp;<em>j&nbsp;&nbsp;</em>(<em>i</em>&nbsp;&le;&nbsp;<em>j</em>) 范围内元素的总和，包含&nbsp;<em>i,&nbsp; j&nbsp;</em>两点。</p>

<p><strong>示例：</strong></p>

<pre>给定 nums = [-2, 0, 3, -5, 2, -1]，求和函数为 sumRange()

sumRange(0, 2) -&gt; 1
sumRange(2, 5) -&gt; -1
sumRange(0, 5) -&gt; -3</pre>

<p><strong>说明:</strong></p>

<ol>
	<li>你可以假设数组不可变。</li>
	<li>会多次调用&nbsp;<em>sumRange</em>&nbsp;方法。</li>
</ol>

 */
public class range_sum_query_immutable {
    

    class Solution {
        public List<String> range_sum_query_immutable(Integer[] param0,Integer[] param1) {
            return null;
        }
    }

    @Test
    public void test(){
        Solution solution = new Solution();
        /**
        ["NumArray","sumRange","sumRange","sumRange"]
[[[-2,0,3,-5,2,-1]],[0,2],[2,5],[0,5]]
        */
        //int [] num1 = new int[]{1,3};
        //int [] num2 = new int[]{2};
        //Assertions.assertEquals(solution.{{questionName}}(num1,num2),2);
    }
}

